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Alex Ar [27]
3 years ago
9

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole o

f zinc reacting for the reaction:Zn(s) + 2HCl(aq) --> ZnCl₂(aq) + H₂(g)
Chemistry
1 answer:
maks197457 [2]3 years ago
5 0

Answer : The enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.

Explanation :

First we have to calculate the moles of Zn and HCl.

\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}

Molar mass of Zn = 65 g/mole

\text{Moles of }Zn=\frac{1.34g}{65g/mole}=0.0206mole

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}=0.750M\times 0.0600=0.0450mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)

From the balanced reaction we conclude that

As, 1 mole of Zn react with 2 mole of HCl

So, 0.0206 moles of Zn react with 0.0206\times 2=0.0412 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and Zn is a limiting reagent and it limits the formation of product.

Now we have to calculate the enthalpy change per mole of zinc reacting for the reaction.

From the reaction we conclude that,

As, 0.0206 moles of Zn produces heat = 3.14 kJ

So, 1 mole of Zn produces heat = \frac{3.14kJ}{0.0206mol}=152.4kJ/mol

Therefore, the enthalpy change per mole of zinc reacting for the reaction is 152.4 kJ/mol.

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